A ship is travelling between two port cities with a speed

­May I please know, how are we considering that , 
\(\frac{\frac{3D}{4}}{x}\)

It is not mentioned in the question explicitly that, ship would have taken the same time as travelling in the velocity 'v', if the ship had started from the 25% of the distance which it had alredy covered?
@chetan2u , can you please explain me, how have you made this assumption. 

Thank you 
 ­

­Let the total distance between the 2 ports be 4D.
Now, 25% of 4D = 1D
Let the speed of the ship at the first half of the journey be x when the ship travelled distance D and then returned because of an emergency. 
If the emergency wouldn't have occurred, the rest of the distance would have been covered at speed x only.
Thus, 3D distance at speed x. 
Time, in this case = 3D/x

Given that the speed after restarting the journey was v
Since, the ship will reach at the same time as before, it means that total distance 4D + D (while going back to the port) = 5D will be covered at speed v
Time, in this case = 5D/v

Both time are equal, so,
3D/x = 5D/v
v/x = 5/3

%= (v-x/x)*100
=(5-3/3)*100
=200/3 = 66.67%, option C